Question

One more question on this test guys. please answer as fast as possible. If you do,

Answer 1

1. The given rectangular equation is $x=2$.

We substitute $x=r\cos \theta$.

$r\cos \theta=2$

Divide through by $\cos \theta$

$r=\frac{2}{\cos \theta}$

$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

$x^2+y^2=36$

This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

This implies that:

$r^2=6^2$

$\therefore r=6$

$\boxed{x^2+y^2=36\to r=6}$

3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

Divide through by r

$r=2\sin \theta$

$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

4. We have $x=\sqrt{3}y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta\sqrt{3}$

This implies that;

$\tan \theta=\frac{\sqrt{3}}{3}$

$\theta=\frac{\pi}{6}$

$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

5. We have $x=y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta$

This implies that;

$\tan \theta=1$

$\theta=\frac{\pi}{4}$

$\boxed{x=y\to \theta=\frac{\pi}{4}}$